Chi sqaure & confidence interval

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SUMMARY

The discussion focuses on calculating the Chi-Squared statistic and confidence interval for a sample of observed frequencies {5, 43, 60, 30, 4}. The Chi-Squared statistic is computed using the formula \(\chi^2=\sum \frac{(O-E)^2}{E}\), where O represents observed frequencies and E represents expected frequencies. For a confidence interval concerning the true variance \(\sigma^2\), the relevant formula is \(P\left( C_{1}< \frac{(n-1)s^2}{\sigma^2} < C_{2} \right) = 1- \alpha = .99\), with specific calculations for C1 and C2 based on the Chi-Squared distribution. The discussion emphasizes the necessity of defining null and alternative hypotheses for conducting the Chi-Squared test.

PREREQUISITES
  • Understanding of Chi-Squared tests and their applications
  • Familiarity with statistical hypotheses (null and alternative)
  • Knowledge of variance and standard deviation concepts
  • Basic proficiency in statistical formulas and calculations
NEXT STEPS
  • Study the Chi-Squared distribution and its properties
  • Learn how to calculate expected frequencies in Chi-Squared tests
  • Explore confidence interval calculations for variance using statistical software
  • Review hypothesis testing methodologies in statistics
USEFUL FOR

This discussion is beneficial for statisticians, data analysts, and students studying probability and statistics, particularly those interested in hypothesis testing and variance analysis.

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Lets say you have 5 trials, and 5 output let's say {5, 43, 60, 30 , 4}...so how would u get the chi sqaure from here & the confidence interval. i haven't work on prob &statc in years, thanks a lot for help folks
 
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I'm supposing you made up this problem... for your random sample of size n=5 is {5, 43, 60, 30 , 4},these are your observed cells/frequencies, but you do not have have your expected cells/frequencies.
 
Basically u can do the confidence for 99%
 
I don't understand what you're trying to do.

For a Chi-Squared test, you need a null hypothesis and alternative hypothesis.
The test statistical value is \chi^2=\sum \frac{(O-E)^2}{E}
where O is a shorthand notation for your observed cells and E is the shorthand notation for expected cell/frequency.

If you are trying to find the confidence interval for the true variance \sigma^2, then the formula for that is

P\left( C_{1}&lt; \frac{(n-1)s^2}{\sigma^2} &lt; C_{2} \right) = 1- \alpha = .99 since 100(1-alpha)% =99%

where C_{1}=\frac{(n-1)s^2}{\chi_{\alpha/2,n-1}} is the lower limit
and
C_2 = \frac{(n-1)s^2}{\chi_{1-\alpha/2,n-1}} is your upper limit.
 
Last edited:
lets say u consider #5 as the thresh#, and for the numbers> 5, u tryto find the confiendnce interval for 99.99% or greater. thanks
 

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